Odpowiedź:
a)
- x² - 5x + 6 = 0
a = - 1 , b = - 5 , c = 6
Δ = b² - 4ac = (- 5)² - 4 * (- 1) * 6 = 25 + 24 = 49
√Δ = √49 = 7
x₁ = ( - b - √Δ)/2a = (5 - 7)/(- 2) = - 2/(- 2) = 2/2 = 1
x₂ = (- b + √Δ)/2a = (5 + 7)/(- 2) = 12/(- 2) = - 6
b)
x² - 5x + 7 = 0
a = 1 , b = - 5 , c = 7
Δ = b² - 4ac = (- 5)² - 4 * 1 * 7 = 25 - 28 = - 3
Δ < 0 więc nie ma miejsc zerowych
x₀ ∈ ∅ (zbiór pusty)
c)
- x² + 10x - 25 = 0
a = - 1 , b = 10 , c = - 25
Δ = b² - 4ac = 10² - 4 * (- 1) * (- 25) = 100 - 100 = 0
x₁ = x₂ = - b/2a = - 10/(- 2) = 10/2 = 5
